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  1. Divisible Sum Pairs
  2. Discussions

Divisible Sum Pairs

  • sawanpatel2508
     + 0 comments

    Typescript solution: I have made the algorithm to run o(n) operation using the hash map, and much better compared to o(n^2) if we use brut force method such as nested for loop. 

    function divisibleSumPairs(n: number, k: number, ar: number[]): number {
  // Write your code here
  const freq: { [key: number]: number } = {};
  let count = 0;
  ar.forEach((int) => {
    const remainder = int % k;
    const complement = (k - remainder) % k;
    if (complement in freq) {
      count += freq[complement];
    }
    if (remainder in freq) {
      freq[remainder]++;
    } else {
      freq[remainder] = 1;
    }
  });
  return count;
}